A396205 a(n) = number of integers k <= n with g(k)^3 < k, where g(k) is the largest divisor d of k satisfying d <= sqrt(k) (A033676).

0, 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22, 22, 22, 23, 24, 24, 24, 24, 25, 25, 26, 26, 26, 26, 27, 28, 29, 29

Offset

1,3

Links

Table of n, a(n) for n=1..60.

Formula

a(n) = Sum_{k=1..n} [ g(k)^3 < k ], where g(k) = A033676(k) and [] denotes the Iverson bracket.

a(n) = Sum_{k=1..n} [ A006530(k) > k^(2/3) ].

For n >= 1, A000720(n) = a(n) - Sum_{c=2..floor(n^(1/3))} ( A000720(floor(n/c)) - A000720(c^2) ).

Example

g(4) = 2 and 2^3 = 8 is not less than 4, so the indicator at k = 4 is 0; hence a(4) = a(3) = 2.
g(5) = 1 and 1^3 = 1 < 5, so the indicator at k = 5 is 1; hence a(5) = 3.
g(10) = 2 (since 2 divides 10 and 2 <= sqrt(10) < 5) and 2^3 = 8 < 10, so the indicator at k = 10 is 1; hence a(10) = 5.

Mathematica

g[n_] := Max[Select[Divisors[n], #^2 <= n &]];
Accumulate[Table[If[g[k]^3 < k, 1, 0], {k, 1, 60}]]

Prog

(Python)
from sympy import divisors
def g(n):
    return max(d for d in divisors(n) if d*d <= n)
def a(n):
    return sum(1 for k in range(1, n+1) if g(k)**3 < k)
print([a(n) for n in range(1, 61)])

Crossrefs

Cf. A033676, A006530, A000720.

Sequence in context: A389719 A194173 A028825 * A384513 A326492 A132924

Adjacent sequences: A396202 A396203 A396204 * A396206 A396207 A396208

Keyword

nonn,easy

Author

Eric Fodge, May 18 2026

Status

approved